On Euler phi function

Solution 1:

We want to prove that if $p$ is such that $2p$ is not in the image of the phi function but $4p$ is then $4p+1$ is prime.

We prove it for $p>5$.

Suppose $\varphi(n)=4p$, notice that $n$ must have a prime factor of the form $kp+1$ in order for $\varphi(n)$ to be a multiple of $p$ while still being less than $4p$. ( because it is impossible to have $p^2 | n$ as $(p-1)p>4p$).

If the prime factor is $2p+1$ we have that $2p$ is in the image of phi.

If the prime factor is $3p+1$ we have $3|\varphi(n)=4p$

If the prime factor is $kp+1$ with $k>4$ we have $\varphi(n)>4p$

So the prime factor is $4p+1$